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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 901. |
Energy per unit volume for a capacitor having area A and separation d kept at potential diffeence V is given by : -A. `(1)/(2)epsilon_(0)(V^(2))/(d^(2))`B. `(1)/(2epsilon_(0))(V^(2))/(d^(2))`C. `(1)/(2)CV^(2)`D. `(Q^(2))/(2C)` |
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Answer» Correct Answer - A Energy density `=(1)/(2)epsilon_(0)E^(2)` Since `E=(V)/(d)` Therefore, energy density `=(1)/(2)epsilon_(0)((V^(2))/(d^(2)))` |
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| 902. |
Two concentric spheres kept in air have radii R and r. They have similar charge and equal surface charge density `sigma`. The electrical potential at their common centre is (where, `epsi_(0) =` permittivity of free space)A. `(sigma(R+r))/(epsilon_(0))`B. `(sigma(R-r))/(epsilon_(0))`C. `(sigma(R+r))/(2epsilon_(0))`D. `(sigma (R+r))/(4epsilon_(0))` |
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Answer» Correct Answer - A Potential at the centre due to sphere 1, `V_(1)=(1)/(4pi epsilon_(0))(q)/(R)` and due to sphere 2 is `V_(2)=(1)/(4pi epsilon_(0))(q)/(r)` `therefore` Electric potential at the common centre `V=V_(1)+V_(2)=(1)/(4pi epsilon_(0))(q)/(R)+(1)/(4pi epsilon_(0))(q)/(r)` `V=(q)/(4pi epsilon_(0))[(1)/(R)+(1)/(r)]=(q)/(4pi epsilon_(0))[(R+r)/(Rr)]` we have `(q)/(4pi Rr)=sigma` `rArr" Electric potential, V"=(sigma(R+r))/(epsilon_(0))` |
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| 903. |
Two identical plane metallic surfaces A and B are kept parallel to each other in air separated by a distance of 1.0 cm as shown in fig. Surface A is given a positive potential of 10V and the outer surface of B is earthed. (i) What is the magnitude and direction of uniform electric field between points Y and Z ? (ii) What is work done in moving a charge of `20 muc` from point X to Y, where X is situated on surface A ? |
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Answer» (i) Magnitude of electric field, `E = (dV)/(dr) = (10V)/(1xx10^(-2)m) = 10^(3) Vm^(-1)` The direction fo electric field is form Y to Z. (ii) As surface A is an equipotenital surface , therefore pot. Diff between X and Y is i..e, `DeltaV = 0` Work done `= q. DeltaV = 0`, i.e, no work is done is moving a charge from X to Y. |
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| 904. |
Give two properties of electric field lines. |
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Answer» 1. Electrons field lines are continous curves. They start from a positively charged body and end at a negatively charged body and end at a negatively charged body. They are continuous but do not form closed loops as there are no lines of force inside the charged body. 2. Tangent to the electric field line at any point gives the directions of electric field intensity at that point. |
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| 905. |
A small particale carying a negative charge of `1.6xx10^(-19)C` ia suspended in equilibrium between the horizontal metal plates 5 cm apart, having a potential difference of 3000 V across them. Find the mass of the particle. |
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Answer» Here, `q = -1.6xx10^(-19)C`. `dr = 5 cm = -5xx10^(-2)m, dV = 3000 V, m = ?` `E = (dV)/(dr) = (-3000)/(5xx10^(-2)) = -6xx10^(4) Vm^(-1)` As the charged particle remains suspended in equalibrium, therefore, `F = mg = qE` or `m = (qE)/(8) = (-1.6xx10^(-19)xx(-6xx10^(4)))/(9.8)` `= 9.8xx10^(-16) kg` |
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| 906. |
A small particle has charge `-5.00muC` and mass `2.00 xx 10-4` kg. It moves from point A where the electric potential is `V_A = +200 V`. to point B, where the electric potential is `V_B=+ 800 V`. The electric force is the only force acting on the particle. The particle has speed `5.00 m/s` at point A. What is its speed at point B? is it moving faster or slower at B than at A. Explain, |
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Answer» Correct Answer - A::B::D `k_A+U_A=k_B+U_B` `:. 1/2mv_A^2+qV_A=1/2mv_B^2+qV_B` `:. v_B=sqrt(v_B^2+2/mq(V_A-V_B)` `=sqrt((5)^2+(2xx(-5xx10^-6))/(2xx10^-4)(200-800)` `=7.42m//s` `v_B-v_A` as the negative charge is moving (freely) from lower potential at `A` to higher potential at `B`. So, its electrostatic `PE` will decrease and kinetic energy will increase. |
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| 907. |
A sphere `S_(1)` of radius `r_(1)` encloses a total charge Q. If there is another concentric sphere `S_(2)` of radius `r_(2) (gt r_(1))` and there be no additional charges between `S_(1) and S_(2)` find the ration of electric flux through `S_(1) and S_(2)`, |
| Answer» `(phi_(1))/(phi_(2)) = (q_(1)// in_(0))/(q_(2)// in_(0)) = (q_(1))/(q_(2)) = (Q)/(Q) = 1` | |
| 908. |
Conducting sphere of radius `R_(1)` is covered by concentric sphere of radius `R_(2)`. Capacity of this combination is proportional toA. `(R_(2)-R_(1))/(R_(1)R_(2))`B. `(R_(2)+R_(1))/(R_(1)R_(2))`C. `(R_(1)R_(2))/(R_(1)+R_(2))`D. `(R_(1)R_(2))/(R_(2)-R_(1))` |
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Answer» Correct Answer - D Capacity of spherical capacitor, `C=4piepsilon_(0)((R_(1)R_(2))/(R_(2)-R_(1))) rArr C prop (R_(1)R_(2))/(R_(2)-R_(1))` |
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| 909. |
A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor.A. Equal and opposite charge will appear in the forces of metal sheetB. Capacity remain sameC. Potential difference between the plates will increaseD. Battery supplies more charge |
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Answer» Correct Answer - B In this case, capacity remains same. |
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| 910. |
Charges `Q_1 and Q_2` lie inside and outside, respectively, of a closed surface S. Let E be the field at any point on S and `phi` be the flux of E over S.A. If `Q_1` changes both E and `phi` will changeB. If `Q_2` changes, will changes but `phi` will not changeC. If `Q_1=0` and `Q_2!=0`, then `E!=0` but `phi=0`D. If `Q_1!=0` and `Q_2=0,` the `E=0` but `phi !=0` |
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Answer» Correct Answer - A::B::C electric fiekld at an point depends on both chares `Q_1` and `Q_2` But electric flux passing from any closed surface depends on the charged enclosed by that closed surface only |
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| 911. |
A neutral pendulum oscillates in a uniform electric field as shown in figure. If a positive charge is given to the pendulum, then its time period A. will increaseB. will decreaseC. will remain constantD. will first increase then decrease |
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Answer» Correct Answer - A `g_e=g-((qE)/m)` or `g_e` will decrease. Hence, `T=2pi sqrt (l/g_e)` will increase |
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| 912. |
Find out following (i) `V_(A)-V_(B)` (ii) `V_(B)-V_(C)` (iii) `V_(C)-V_(A)` (iv) `V_(D)-V_(C)` (v) `V_(A)-V_(D)` (vi) Arrange the order of potential for points A, B, C and D. |
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Answer» (i) `|DeltaV_(AB)|=Ed=20xx2xx10^(-2)=0.4` So, `V_(A)-V_(B)=0.4 V` because in the direction of electric field potential always decreases. (ii) `|DeltaV_(BC)|=Ed=20xx2xx10^(-2)=0.4` so, `V_(B)-V_(C)=0.4 V` (iii) `|DeltaV_(CA)|=Ed=20xx4xx10^(-2)=0.8` so, `V_(C)-V_(A)=-0.8 V` because In the direction of electric field potential always decreases. (iv) `|DeltaV_(DC)|=Ed=20xx0=0` so, `V_(D)-V_(C)=0` because the effective distance D and C is zero. (v) `|DeltaV_(AD)|=Ed=20xx4xx10^(-2)=0.8` so, `V_(A)-V_(D)=0.8 V` because In the direction of electric field potential always decreases. (vi) The order of potential is : `V_(A) gt V_(B) gt V_(C)=V_(D)`. |
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| 913. |
Two point charges `Q_1` and `Q_2` lie along a line at a distance from each other. Figure 3.7 shows the potential variation along the line of charges. At which of the points 1, 2, and 3 is the electric field zero ? What are the singns of the charges `Q_1` and `Q_2` and which of the two charges is greater in magnitude ? . |
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Answer» Correct Answer - (a). `q_(1)tove;q_(2)to-ve` `q_(1)` has larger magnitude (b). `(d)/(sqrt(3)-1)` |
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| 914. |
Identify the correct statement about the charges `q_1` and `q_2` then A. `q_1` and `q_2` both are positiveB. `q_1` and `q_2` both are negativeC. `q_1` is positive `q_2` negativeD. `q_2` is positive and `q_1` is negative |
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Answer» Correct Answer - B electric lines terminate on negative charge |
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| 915. |
Units of electric flux areA. `(N-m^2)/C^2`B. `N/(C^2-m^2)`C. volt-mD. volt-m^3 |
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Answer» Correct Answer - C `phi=ESrarr (V/m)(m^2)rarr"volt"-m` |
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| 916. |
The electrical potential function for an electrical field directed parallel to the x-axis is shown in the given graph. Draw the graph of electric field strength. |
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Answer» Correct Answer - A::B::C::D `E=-(dV)/(dx)=-` Slope V-x graph. From `x=-2m` to `x=0`, slope `=+5V//m` `:. E=-5V//m` From `x=0` to `x=2m,` slope `=0` `:.E=0` From `x=2m` to x=4m`, slope `=+5V//m,` `:. E=-5V//m` From `x=4m` to x=8m`, slope `x=--5V//m` `:. E=+5V//m` Corresponding E-x graph is as shown in answer. |
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| 917. |
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 x 105 N C -1. It experiences a torque equal to 8 N m. The charge on the dipole if the dipole length is 1 cm is- (a) 4 mC (b) 8 mC (c) 5 mC (d) 1 mC |
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Answer» correct answer is (b) 8 mC |
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| 918. |
A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness `d_(1)` and dielectric constant `K_(1)` and the other has thickness `d_(2)` and dielectric constant `K_(2)` as shown in figure. This arrangement can be through as a dielectric slab of thickness `d (= d_(1) + d_(2))` and effective dielectric constant `K`. The `K` is. .A. `(k_(1) d_(1) + k_(2) d_(2))/(d_(1) + d_(2))`B. `(k_(1) d_(1) + k_(2) d_(2))/(k_(1) + k_(2))`C. `(k_(1) k_(2) (d_(1) + d_(2)))/((k_(1) d_(2) + k_(2) d_(1)))`D. `(2 k_(1) k_(2))/(k_(1) + k_(2))` |
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Answer» Correct Answer - C The capacities of two individual condensers are `C_(1) = (k_(1) in_(0) A)/(d_(1))` and `C_(2) = (k_(2) in_(0) A)/(d_(2))` The arrangement is equivalent to two capacitors joined in series. Therefore, the combined capacity `(C_(s))` is given by `(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) = (d_(1))/(k_(1) in_(0) A) + (d_(2))/(k_(2) in_(0) A) = (1)/(in_(0) A) [(d_(1))/(k_(1)) + (d_(2))/(k_(2))] = (d_(1) k_(2) + d_(2) k_(1))/(in_(0) A K_(1) K_(2))` or `C_(s) = (in_(0) A k_(1) k_(2))/(d_(1) k_(2) + d_(2) k_(1)) = (k in_(0) A)/((d_(1) + d_(2))) :. K = (k_(1) k_(2) (d_(1) + d_(2)))/((d_(1) k_(2) + d_(2) k_(1)))` |
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| 919. |
Consider three identical metal spheres ` A, B` and C. Spheres A carres charge ` + 6q` carries no charge . Spheres A and B are touchd together and them separated. Sphere C is then touched to sphere `A` and separated from it . Finally the sphere `C` is touched to sphere `B` and separated from it . Find the final chage on the sphere `C`. |
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Answer» Correct Answer - ` 1. 125 Q` When sphere `A` and `B` are touched and seperated `q_A = q _B = (+_ 6 Q - 3 q) /2` `q-A = q _B = (3q)/2` when sphere `A` and `C` are toched and seperated `q_A =q_C = (3 Q //2 +0)/2 = (3q)/4` Now `B` and `C` are touched and seperated `q_B =q_C = (3 q// 2 + 3 q //4)/2 = (9q)/8 = 1.12q`. |
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| 920. |
In a certain `0.1 m^(3)` of space, electric potential is found to be 5 V throughout. What is the electric field in this region ? |
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Answer» As `E = - (dV)/(dr)` and V is constant throughout, therefore, `E = 0` |
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| 921. |
Suppose that the earth has a net charge that is not zero. Is it still possible to adapt the earth as a standard reference point if potential and assign the potental `V = 0 to it? |
| Answer» Yes, earth can still be taken as a standard reference point for potential V = 0, because capacity of earth is infinitely large. | |
| 922. |
Will there be any effect on potential at a point if the medium around this point is changed ? |
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Answer» Yes, Potential at a point decreases when dielectric constant (K) of the medium increases. This is because `V = q//4 pi in_(0)` Kr. |
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| 923. |
How does a free electrons at rest move in an electric filed. |
| Answer» A free electron at rest will move in a direction opposire to electric field. | |
| 924. |
Write the dimensional formula of electric field.S |
| Answer» `[MLT^(-3) A^(-3)] ` | |
| 925. |
Name any four vector fields. |
| Answer» Electric field, maganetic field gravitational field and flow field(of a liquid) are vector fields. | |
| 926. |
A metal foil of negligible thickness is introduced between the two plates of a capacitor at the centre. The capacitance of capacitor will beA. sameB. doubleC. halfD. k times |
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Answer» Correct Answer - A `C=(in_(0)A)/(d-1(1-1/k))` For metal `k=infty` and t is negligible. `therefore" "C=(in_(0)A)/d="same"` |
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| 927. |
A proton is placed in a unifrom electric field directed along the positive x-axis. In which direction willl it tend to move? |
| Answer» As proton is positively charged, it will tend to move along + x axis,i..e., along the direction of electric field. | |
| 928. |
When air is replaced by a dielectric medium of constant K, the maximum force separated by a distanceA. increases k timesB. remains unchangedC. increases `k^(2)` timesD. decreases k times |
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Answer» Correct Answer - A `C_(m)=C_("air")k=Ck="Increases by k times"` |
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| 929. |
The condenser always block D. C. permitsA. pulsating current through itB. alternating current through external circuitC. only half cycle of alternating current through itD. none of these |
| Answer» Correct Answer - B | |
| 930. |
Electric potential at any point in equatorial plane of a dipole is ………. . |
| Answer» Electric potential, V = 0. | |
| 931. |
Do electrons tend to go to regions of high potential ? |
| Answer» As electrons are negatively charged, they have tendency to go to regions of high potential. | |
| 932. |
An arrangement which consist of two conductors separated by a dielectric medium is calledA. resistorB. inductorC. rectifierD. capacitor |
| Answer» Correct Answer - D | |
| 933. |
The ohmic resistance of a condenser for D.C. current isA. finiteB. infiniteC. zeroD. `V//1` |
| Answer» Correct Answer - B | |
| 934. |
The middle point fo a conductor is earthed and its ends are maintained at a potential at the two ends at the middle point? |
| Answer» Potential at the middle point is zero. The values of potential at the two ends of the conductor must be `+110V and -110V`, so that pot. Diff, across the two ends is `110 - (-110) = 220V`. | |
| 935. |
The earthed conductor plate of a condenser helps toA. decrease the potential of charged conductorB. increase the potential of charged conductorC. keep constant the potential of charged conductorD. none of these |
| Answer» Correct Answer - A | |
| 936. |
Assertion Due to two charges electric field cannot be zero at some simultaneously. Reason Field is a vector quantityA. If both Assertion and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are correct but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true |
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Answer» Correct Answer - B |
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| 937. |
`A,B` and `C` are three concentric metallic shells . Shell `A` is the innermost and shell `C` is the outermost. `A` is given some charge (i) The inner surfaces of `B`and `C` will have the same charge (ii) The inner surfaces of `B` and `C` will have the same charge density (iii) The outer surfaces of `A,B` and `C` will have the same charge (iv) The outer surfaces of `A,B` and `C` will have the same charge densityA. The inner surfaces of `B` and `C` will have the same charge.B. The iner surface of `B` and `C` will be the same charge density.C. The outer surfaces of `A,B` and `C` will have the same charge.D. The outer surfaces of `A,B` and `C` will have the same charge density. |
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Answer» Correct Answer - A::B::D |
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| 938. |
A parallel plate capacitor is charged from a cell and then isolated from it. The separation between the plates is now increasedA. The force of attraction between the plates will decrease.B. The field in the region between the plates will not change.C. The energy stored in the capacitor will increase.D. The potential difference between the plates will decrease. |
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Answer» Correct Answer - B::C |
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| 939. |
A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential, capacitance respectively are (A) Constant, decreases, decreases (B) Increases, decreases, decreases (C) Constant, decreases, increases (D) Constant, increases, decreases |
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Answer» Correct option is: (A) Constant, decreases, decreases |
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| 940. |
A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is(A) \(C=\cfrac{Aε_0}{d}\left(\cfrac{k+3}{4k}\right)\)(B) \(C=\cfrac{Aε_0}{d}\left(\cfrac{2k}{k+3}\right)\)(C) \(C=\cfrac{Aε_0}{d}\left(\cfrac{k+3}{2k}\right)\)(D) \(C=\cfrac{Aε_0}{d}\left(\cfrac{4k}{k+3}\right)\) |
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Answer» Correct option is: (D) \(C=\cfrac{Aε_0}{d}\left(\cfrac{4k}{k+3}\right)\) |
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| 941. |
Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio.(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 1 : 3 |
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Answer» Correct option is: (A) 1 : 1 |
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| 942. |
Charge + q and -q are placed at points A and B respectively which are distance 2L apart. C is the mid point of A and B. The work done in moving a charge +Q along the semicircle CRD as shown in the figure below is(A) \(\cfrac{-qQ}{6\piε_0L}\)(B) \(\cfrac{qQ}{2\piε_0L}\)(C) \(\cfrac{qQ}{6\piε_0L}\)(D) \(\cfrac{-qQ}{2\piε_0L}\) |
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Answer» Correct option is: (A) \(\cfrac{-qQ}{6\piε_0L}\) |
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| 943. |
A parallel plate capacitor has circular plates of radius 8 cm and plate separation 1mm. What will be the charge on the plates if a potential difference of 100 V is applied?(A) 1.78 × 10-8 C (B) 1.78 × 10-5 C (C) 4.3 × 104 C (D) 2 × 10-9 C |
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Answer» Correct option is: (A) 1.78 × 10-8 C |
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| 944. |
If the difference between the radii of the two spheres of a spherical capacitor is increased, state whether the capacitance will increase or decrease. |
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Answer» The capacitance of a spherical capacitor is C = 4πε0\(\left(\cfrac{ab}{b-a}\right)\) where a and b are the radii of the concentric inner and outer conducting shells. Hence, the capacitance decreases if the difference b – a is increased. |
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| 945. |
A charge q is moved from a point A above a dipole of dipole moment p to a point B below the dipole in equitorial plane without acceleration. Find the work done in this process. |
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Answer» The equatorial plane of an electric dipole is an equipotential with V = 0. Therefore, the no work is done in moving a charge between two points in the equatorial plane of a dipole. |
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| 946. |
Assertion: A small electric dipole is moved translatonally from higher potential to lower potential in uniform electric field. Work done by electric is positive. Reason: When a positive charge is moved from higher potential to lower potential, work done by electric field positive.A. If both Assertion and Reason are true but Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D In uniform electrilc field net force on an electric dipole =0 therefore no work is done in transaltaional motion of the dipole. Electric lines also flow from higher potential to lower potential. Electrostatical force of positive charge acts in the direction of electric field. Therefore, work done is positive. |
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| 947. |
Assertion: In the electric field `E=(4eci+4hatj)N//C` electric potential at A (4m, 0) is more than the electric portential at `B(0,4m)` Reason: Electric lines of forces always travel from higher potential to lower potential.A. If both Assertion and Reason are true but Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D `V_A=V_b=-int_B^A E.dr` `=-int_((0,4))^((4,0)) (4hati+4hatj).(dxhati+dyhatj)` `=-int_((0,4))^((4,0)) (4x+4dy)=0` `:. V_A=V_b` |
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| 948. |
Assertion: Two charges `-q` each are fixed points A and B. When a third chrge `-q` is moves from A to B, electrical potential energy first decreases than increases, Reason: Along the line joining A and B, third charge is stable equilibrium position at centre.A. If both Assertion and Reason are true but Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A::B At stable equililbrim position, potential energy is minimum. |
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| 949. |
A particle of mass 2 kg chrge 1 mC is projected vertially with velocity k`10 ms^-1`. There is as uniform horizontal electric field of `10^4N//C,` thenA. the horizontal range of the particle is `10 m`B. the time of flight of the particle is `2s`C. the maximum heighty reached is `5m`D. the horizontal range of the particle is `5m` |
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Answer» Correct Answer - A::B::C `T=(2u_y)/g=(2xx10)/10=2s` `H=u_y^2/(2g)=((10)^2)/20=5m` `R=1/2a_xT^2=1/2((qE)/m)T^2` `=1/2((10^-3xx10^4)/2)(2)^2` `=10m` |
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| 950. |
A charge ` + 10^-9 C` is located at the origin in free space `&` another charge ` Q` at ` (2,0,0)` . If the X-component of the electric field at (3,1,1) is zero , calculate the value of `Q`, Is the Y-component zero at (3,1,1) ? |
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Answer» Correct Answer - `-0.46xx10^(-9)C` |
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